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I'm doing Leonard Susskind's course on General Relativity (http://deimos3.apple.com/WebObjects/Core.woa/Feed/itunes.stanford.edu-dz.19344853322.019344853324 ), and I'm stuck on a particular derivation (in Lecture 4, about 12 minutes in). He left as an exercise to calculate: $D_m V^n$. He gave us the answer already as:

    $D_m V^n=\partial_m V^n+\Gamma _{\text{mr}}^nV^r $

I can do most of the derivation as follows:

    $D_m V^n=D_m\left(g^{\text{np}} V_p\right)$

    $D_m V^n=\left(D_m g^{\text{np}}\right)V_p - g^{\text{np}}\left(D_m V_p\right) $     Product rule

    $D_m V^n=g^{\text{np}}\left(D_m V_p\right)$                            $D_m g^{\text{np}}=0$ in Gaussian-normal coordinates

    $D_m V^n=g^{\text{np}}\left(\partial_m V_p-V_r \Gamma _{\text{mp}}^r\right)$            apply covariant derivative

    $D_m V^n=\partial_m g^{\text{np}}V_p -g^{\text{np}} V_r \Gamma _{\text{mp}}^r$           multiply through

    $D_m V^n=\partial_m V^n-g^{\text{np}} \Gamma _{\text{mp}}^rV_r$

This last step is where I am stuck. How to get from
$$-g^{\text{np}} \Gamma _{\text{mp}}^rV_r$$ to $$\Gamma _{\text{mr}}^nV^r $$ I have some idea that the minus sign will be taken care of by $g^{\text{np}}=-g_{\text{np}}$, and there must be some swap of summation indices, but I can't seem to see it.

Extropy
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1 Answers1

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In the step "multiply through", you commuted the metric with the partial derivative - you should have gotten an extra term involving the partial derivative of the metric here. If you then expand $\Gamma$ in terms of the partial derivatives of the metric you find that this new term ends up leaving you with $\Gamma$ with some different indices and a flipped sign.

That's an unnecessary computation, however - you can see this fact without expanding the $\Gamma$ from the product rule: $dx^i (D_j \partial_k) = - (D_j dx^i) \partial_k = \Gamma^{i}_{jk}$. I'm not sure what formalism/notation Susskind covers; this may make more sense to you written out as $$ \begin{align} Y_i D_j X^i &= D_j (Y_i X^i) - X^i D_j Y_i \\ &=\partial_j(Y_i X^i) - X^i (\partial_j Y_i - \Gamma_{ji}^k Y_k) \\ &=Y_i\partial_jX^i+X^i\Gamma_{ji}^kY_k \\ &=Y_i (\partial_j X^i + \Gamma^i_{jk}X^k). \end{align} $$

Since this is true for any $Y$ you can just "divide out" by $Y_i$.

Anthony Carapetis
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